python将多个排序列表逐个组合成一个大的排序列表

Question

我有几个排序列表,我想将它们一起添加到一个大的排序列表中.最有效的方法是什么？

这是我要做的,但效率太低:

big_list=[]
for slist in sorted_lists: # sorted_lists is a generator, so lists have to be added one by one
    big_list.extend(slist)
big_list.sort()

以下是sorted_lists的示例:

sorted_lists的大小= 200

sorted_lists的第一个元素的大小= 1668

sorted_lists=[
['000008.htm_181_0040_0009', '000008.htm_181_0040_0037', '000008.htm_201_0041_0031', '000008.htm_213_0029_0004', '000008.htm_263_0015_0011', '000018.htm_116_0071_0002', '000018.htm_147_0046_0002', '000018.htm_153_0038_0015', '000018.htm_160_0060_0001', '000018.htm_205_0016_0002', '000031.htm_4_0003_0001', '000032.htm_4_0003_0001', '000065.htm_5_0013_0005', '000065.htm_8_0008_0006', '000065.htm_14_0038_0036', '000065.htm_127_0016_0006', '000065.htm_168_0111_0056', '000072.htm_97_0016_0012', '000072.htm_175_0028_0020', '000072.htm_188_0035_0004'….],
['000018.htm_68_0039_0030', '000018.htm_173_0038_0029', '000018.htm_179_0042_0040', '000018.htm_180_0054_0021', '000018.htm_180_0054_0031', '000018.htm_182_0025_0023', '000018.htm_191_0041_0010', '000065.htm_5_0013_0007', '000072.htm_11_0008_0002', '000072.htm_14_0015_0002', '000072.htm_75_0040_0021', '000079.htm_11_0005_0000', '000079.htm_14_0006_0000', '000079.htm_16_0054_0006', '000079.htm_61_0018_0012', '000079.htm_154_0027_0011', '000086.htm_8_0003_0000', '000086.htm_9_0030_0005', '000086.htm_11_0038_0004', '000086.htm_34_0031_0024'….],
['000001.htm_13_0037_0004', '000008.htm_48_0025_0006', '000008.htm_68_0025_0008', '000008.htm_73_0024_0014', '000008.htm_122_0034_0026', '000008.htm_124_0016_0005', '000008.htm_144_0046_0030', '000059.htm_99_0022_0012', '000065.htm_69_0045_0017', '000065.htm_383_0026_0020', '000072.htm_164_0030_0002', '000079.htm_122_0030_0009', '000079.htm_123_0049_0015', '000086.htm_13_0037_0004', '000109.htm_71_0054_0029', '000109.htm_73_0035_0005', '000109.htm_75_0018_0004', '000109.htm_76_0027_0013', '000109.htm_101_0030_0008', '000109.htm_134_0036_0030']]

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编辑

谢谢你的答案.我想我应该更明确地说我没有模拟排序列表但我正在迭代一些大文件来获取它们.所以,我需要逐个添加它们,正如我在上面的粗略代码中所示.

Answer 1

标准库heapq.merge为此目的提供:

>>> a=[1,3,5,6]
>>> b=[2,4,6,8]
>>> c=[2.5,4.5]
>>> list(heapq.merge(a,b,c))
[1, 2, 2.5, 3, 4, 4.5, 5, 6, 6, 8]
>>> 

或者,在您的情况下:

big_list = list(heapq.merge(*sorted_lists))

请注意,您不必创建列表,因为heapq.merge返回一个iterable:

for item in heapq.merge(*sorted_lists):

引用文档:

类似sorted(itertools.chain(*iterables))但返回一个iterable,不会同时将数据全部拉入内存,并假设每个输入流已经排序(从最小到最大).